Example 20
Find the derivative of f(x) from the first principle, where f(x) is
(ii) x sin x
Given f (x) = x sin x
We need to find Derivative of f(x)
We know that
fβ(x) = limβ¬(hβ0) πβ‘γ(π₯ + β) β π(π₯)γ/β
Here, f (x) = x sin x
So, f (x + h) = (x + h) sin (x + h)
Putting values
fβ(x) =limβ¬(hβ0) ((π₯ + β) sinβ‘γ (π₯ + β) β π₯ sinβ‘γπ₯ γ γ)/β
Using sin (A + B) = sin A cos B + cos A sin B
= limβ¬(hβ0)β‘γ((π₯ + β)(sinβ‘γπ₯ cosβ‘γβ +γ cos π₯γβ‘sinβ‘γβ γ )γ β π₯ sinβ‘π₯ γ)/βγ
= limβ¬(hβ0)β‘γ(π₯(sinβ‘γπ₯ cosβ‘γβ +γ cosγβ‘γπ₯ sinβ‘γβ) + β (sinβ‘γπ₯ cosβ‘γβ +γ cosγβ‘γπ₯ sinβ‘γβ) β π₯ sinβ‘π₯ γ γ γ γ γ γ γ γ)/βγ
= limβ¬(hβ0)β‘γ(π₯ sinβ‘γπ₯ cosβ‘γβ + π₯ cosβ‘γπ₯ sinβ‘γβ + β γ(sinγβ‘γπ₯ cosβ‘γβ + cosβ‘γπ₯ sinβ‘γβ) β π₯ sinβ‘π₯ γ γ γ γ γ γ γ γ)/βγ
= limβ¬(hβ0)β‘γ(π₯π ππ π₯ cosβ‘γβ β π₯ sinβ‘γπ₯ + π₯ cosβ‘γπ₯ sinβ‘γβ + β(sinβ‘γπ₯ cosβ‘γβ + cosβ‘γπ₯ sinβ‘β γ γ)γ γ γ γ γ)/βγ
= limβ¬(hβ0)β‘γ(π₯π ππ π₯ γ(cosγβ‘γβ β 1)+ π₯ cosβ‘γπ₯ sinβ‘γβ + β(sinβ‘γπ₯ cosβ‘γβ + cosβ‘γπ₯ sinβ‘β γ γ)γ γ γ γ)/βγ
= limβ¬(hβ0)β‘((π₯ sinβ‘γπ₯ (cosβ‘γβ β1)γ γ)/β+(π₯ cosβ‘γπ₯ sinβ‘β γ)/β+(β (sinβ‘γπ₯ cosβ‘γβ +cosβ‘γπ₯ sinβ‘γβ)γ γ γ γ)/β)
= limβ¬(hβ0)β‘γγx sinγβ‘γπ₯ (cosβ‘γβ β1)γ γ/β+limβ¬(hβ0) (π₯ cosβ‘γπ₯ sinβ‘β γ )/h+limβ¬(hβ0) (sinβ‘γπ₯ cosβ‘γβ +cosβ‘γπ₯ sinβ‘γβ )γ γ γ γ γ
= β x sin x (π₯π’π¦)β¬(π‘βπ) ((πβγ πππγβ‘γπ)γ)/π+π₯ cosβ‘γπ₯ (π₯π’π¦)β¬(π‘βπ) πππβ‘π/π‘+γ limβ¬(hβ0) (sinβ‘γπ₯ cosβ‘γβ +cosβ‘γπ₯ sinβ‘γβ )γ γ γ γ
= β x sin x (0) + x cos x (1) + ( sin x cos 0 + cos x sin 0)
= 0 + x cos x + sin Γ 1 + cos x Γ 0
= 0 + x cos x + sin x + 0
= x cos x + sin x
Hence fβ (x) = x cos x + sin x
Made by
Davneet Singh
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo
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